Chemistry of Natural Products

SEMESTER FINAL EXAM MATERIALS CHEMISTRY CULTURE

Name                          : Putri Mutiara Ishak

NIM                            : RSA1C110002

Course                        : Chemistry of Natural Materials

Credits                        : 2

Lecturer                     : Dr. Syamsurizal, M.Si

Time                           : 22-29 December 2012

1.    Explain the triterpenoid biosynthetic pathway, identify important factors that determine the quantities produced many triterpenoids.

Answer:

               The mechanism of the reaction steps of the biosynthesis of triterpenoids is acetic acid which has been activated by coenzyme A condensing type Claisen asetoasetat produce acid. Compounds resulting from condensation is doing aldol condensation with acetyl coenzyme A, which produces branched carbon chains such as those found in the mevalonic acid stage, next is the phosphorylation reaction, elimination of phosphoric acid and decarboxylation produces isopentenyl (IPP), which later became dimetilalilpiropospat berisomerisasi (DMAPP) by isomeriasi enzyme. IPP as isoprene unti actively join DMAPP is the first step of the polymerization of isoprene to produce triterpenoid. This merger occurs because electrons attack the double bond carbon atoms of IPP to DMAPP followed by electron-deficient pyrophosphate ion discharge that produces geranil. pyrophosphate (GPP). Subsequent amalgamation between the IPP and GPP by the same mechanism produces Farnesil pyrophosphate (FPP). Triterpenoid dibiosintesis of 6 isoprene units, consisting of a C30 acyclic which are precursors of squalen.

2.    Describe the structure determination of flavonoids, specificity and intensity of absorption signal by using IR and NMR spectra. Give the example of at least two different structures.

Answer:
a. 

Infrared spectrophotometric analysis aimed to determine the functional groups of a compound based on the absorption of the electromagnetic spectrum IR region. IR spectra analysis results indicate that catechins were isolated containing functional groups with an approximate functional group C = C aromatic uptake by region 1500-1600 cm-1, the OH group at 2000-3600 catchment area (width). Vibration is used for the identification was bending vibrations, especially rocking vibration (wobble), which is located in the wavenumber region 2000-400 cm-1.

The spectrum of 13C-NMR chemical shift data and the isolated catechins were measured using methanol-D3 with a frequency of 500 MHz. From the data 13CNMR known that the isolated catechins have 15 signals that indicate the presence of as many as 15 pieces of carbon atoms, ie δc 157.8 (C-9), 157.6 (C-7), 156.9 (C-5), 146.28 (C-4 '), 146, 26 (C-3'), 132.2 (C-1 '), 120.1 (C-6'), 116.2 (C-5 '), 115.3 (C-2 '), 100.9 (C-10), 96.3 (C-6), 95.5 (C-8), 82.8 (C-2), 68.8 (C-3), 28.5 (C-4) ppm. The number of carbon atoms is equal to the number of carbon atoms.The results of standard catechin compounds 1H-NMR spectrum of the isolated catechins were measured using methanol-D3 with a frequency of 500 MHz. Chemical shift that occurs at 2.52 (1H, dd), 2.84 (1H, dd), 3.98 (1H, m), 4.57 (1H, d), 5.86 (1H, d), 5.93 (1H, d), 6.72 (1H, dd), 6.76 (1H, d), 6.84 (1H, d). 

b.    IR spectrum shows absorption bands were widened in wavenumber, υ max: 3453 cm-1, indicating a hydroxyl group, absorption band at wavenumber, υ max: 2930 cm-1, an absorption of aliphatic CH, absorption bands at wave numbers, υ max: 1691 cm-1, is the absorption of the C = O of carboxylic acids reinforced by the presence of absorption bands in the wave number, υ max: 1114 cm-1, absorption band at wavenumber, υ max: 1454 cm-1 and CH 2 absorption band at wavenumber, υ max: 1384 cm-1 is the absorption of CH bending of geminal dimethyl that are characteristic triterpenoid compounds that support infrared spectroscopic data.

To determine the number of protons and proton type of the isolated compounds performed proton 1H-NMR spectroscopy using CDCl3 solvent at a frequency 600 MHz. From the 1H-NMR spectrum of the isolated triterpenoid compound known that there are forty-six signal. Each signal is distributed in the proton CH3 and CH2 number forty-one CH proton signal and the third signal. COOH group of the proton and the proton signal alkene HC = C <there is a signal. Furthermore, in order to more clearly signal analysis is carried out against the expansion chemical shift region from 0.8389 to 2.5145 ppm. Expansion of the 1H-NMR spectrum in the region chemical shifts from 0.8389 to 2.5145 ppm, it is known that these compounds have a bunch protons in the region δ: 0.80 to 1.54 ppm which is the chemical shift of the proton CH2 and CH3 protons or groups sheilding (protected) ie protons of hydrocarbon saturated. Seven singlet methyl proton signal (s, 3H), which appears on δH 0.80; 0.84; 0.87; 1.03; 1.00; 1.05; 1.06 ppm, respectively for the H-27, H-26, H-25, H-23, H-24, H-28 and H-30, the signal of carbon DEPT NMR data support the stated that the isolated compounds have the 7 carbon methyl (CH3). Furthermore, the proton signal HC = C <appeared at chemical shift 5.71 ppm (t, 1H), corresponding to the proton signal H-12, This signal also supports DEPT NMR carbon data stating that the compound results insulation has a CH of alkene carbon estimated carbon 12 (C-12). Furthermore, this compound also has a bunch of protons in the region δ: 1.73 to 2.51 ppm which is the chemical shift of the CH proton.

c. Synthesis of 2-hidroksikalkon using IR spectroscopy

The results of the analysis of the synthesis of 2 - hidroksikalkon IR spectroscopy showed a hydroxyl group (OH) with a wide absorption peak at 3455.23 cm-1, carbonyl group (C = O) in the area of ​​1645.99 cm-1, C = C alkene in the area 1598.85 cm-1, C = C aromatic at 1584.09 cm​​-1 region, substituents at the ortho position on the aromatic ring in the area of ​​731.75 cm-1, C phenol at 1150.65 cm-1 region. Uptake of aromatic and alkene = CH above 3000 cm-1 is too weak to be read in the IR spectrum of compounds synthesized.

Synthesis of 2-hidroksikalkon using NMR spectroscopy

 

Analysis of compounds using 1H-NMR spectroscopy can provide information on the number, nature and the environment of hydrogen atoms in a molecule. 1H NMR spectrum of compounds synthesized are presented in the following figure. 
1H NMR spectrum of compounds synthesized absorption-absorption appears at 8.09 ppm region ((1H, d, 6.7 Hz), 6.97 ppm (1H, dd, 7.35 Hz), 6.53 ppm (1H, t, 7.95 Hz), 8.09 ppm (1H, d, 8.55 Hz) indicating the protons of the aromatic ring B. Absorption at 8.28 ppm (1H, d, 15.25 Hz) and 8 , 03 ppm (1H, d, 15.25 Hz) indicates the protons in the-CH = ethylene trans position. absorption-absorption which appeared in the region 7.51 ppm (2H, d, 7.35 Hz), and 7 , 56 ppm (2H, d, 7.65 Hz) and 7.08 ppm (1 H, t, 6.7 Hz) showed the protons on the aromatic ring A.

3.    In the isolation of alkaloids, in the early stages of acid or base required conditions. Explain the basis of the use of reagents, and give examples of at least three kinds of alkaloids.

Answer:

               In the very early stages of alkaloid isolation requires acidic conditions, with the addition of an organic acid extract will produce salt. The addition of acid is also useful for bind alkaloids. The addition of alkaline salts useful for freeing ties into alkaloids.

For example:

a.       Isolation alkaloid made ​​by the method of extraction. the leaves are often high in fat, very nonpolar wax. The compounds are separated from plant material as an initial step in a way melatkan sample with diethylether. Most alkaloids are not soluble in diethylether. However, the extract should always be checked for the presence of alkaloids by using one reagent precipitating alkaloids. When a number of alkaloids soluble in diethylether solvent, the plant material is added to the acid to bind the alkaloid salts.

b.      Working principle is the principle keller alkaloids contained in a will as a form of salt and salt-free into the free alkaloids. For that added a strong base alkaloids. Bases used should not be too strong because alkaloids are generally unstable. At high pH alkoloid will decompose, especially in the free state. Plant material can be released with sodium carbonate.

c. Nicotine can be purified by steam distillation from the basified solution. Solution in water that is acidic and contains alkaloids and alkaloid diekstraksim can basified with an organic solvent, so that neutral and acidic compounds are readily soluble in water left in the water.

4.    Describe the relationship between biosynthesis, methods of isolation and structural determination of compounds of natural ingredients. Give an example.

Answer:

               Biosynthesis, isolation and structural determination of the method of natural materials closely related compounds. In general, the chemical nature of materials related to the isolation, structure determination, and synthesis of organic compounds derived from biological sources. With biosynthesis done, it will be known how the process of formation of a chemical compound, natural ingredients. Then isolation, essentially chemical isolation from natural materials is a process of how we can separate an organic compound in a sample using a suitable solvent. Plants contain thousands of compounds that are categorized as primary metabolites and secondary metabolites. Usually the isolation of compounds from natural materials to isolate the secondary metabolites, secondary metabolites due believed and has been researched to provide benefits to human life. Then be identified through existing spectrum, the spectrum can be determined from the structure. Thus, from the results of the determination of the structure can be compared with the structure of the resulting chemical senywa the biosynthesis process to match the structure.

For example:

biosynthesis of cholesterol

Biosynthesis of cholesterol is directly regulated by the cholesterol levels present, though the homeostatic mechanisms involved are only partly understood. A higher intake from food leads to a net decrease in endogenous production, whereas lower intake from food has the opposite effect. The main regulatory mechanism is the sensing of intracellular cholesterol in the endoplasmic reticulum by the protein SREBP (sterol regulatory element-binding protein 1 and 2). In the presence of cholesterol, SREBP is bound to two other proteins: SCAP (SREBP-cleavage-activating protein) and Insig1. When cholesterol levels fall, Insig-1 dissociates from the SREBP-SCAP complex, allowing the complex to migrate to the Golgi apparatus, where SREBP is cleaved by S1P and S2P (site-1 and -2 protease), two enzymes that are activated by SCAP when cholesterol levels are low. The cleaved SREBP then migrates to the nucleus and acts as a transcription factor to bind to the sterol regulatory element (SRE), which stimulates the transcription of many genes. Among these are the low-density lipoprotein (LDL) receptor and HMG-CoA reductase. The former scavenges circulating LDL from the bloodstream, whereas HMG-CoA reductase leads to an increase of endogenous production of cholesterol. A large part of this signaling pathway was clarified by Dr. Michael S. Brown and Dr. Joseph L. Goldstein in the 1970s. In 1985, they received the Nobel Prize in Physiology or Medicine for their work. Their subsequent work shows how the SREBP pathway regulates expression of many genes that control lipid formation and metabolism and body fuel allocation.

Cholesterol synthesis can be turned off when cholesterol levels are high, as well. HMG CoA reductase contains both a cytosolic domain (responsible for its catalytic function) and a membrane domain. The membrane domain functions to sense signals for its degradation. Increasing concentrations of cholesterol (and other sterols) cause a change in this domain's oligomerization state, which makes it more susceptible to destruction by the proteosome. This enzyme's activity can also be reduced by phosphorylation by an AMP-activated protein kinase. Because this kinase is activated by AMP, which is produced when ATP is hydrolyzed, it follows that cholesterol synthesis is halted when ATP levels are low.

Isolation of Cholesterol from Egg Yolk

               Procedure: In a 250 mL round bottom flask, combine a hard-boiled egg yolk, 1 g of K2CO3, 5 g of sand, and 10 mL of MeOH. Grind together until it is smooth – it will look like soft scrambled eggs. Add 20 mL of cyclohexane, stir thoroughly – it will look like corn meal mush – then warm to reflux. Rotovap off the solvent. While the solvent is rotovapping, prepare a chromatography column with 15 g of flash silica gel and a layer of sand on top, and have fifteen test tubes ready to take fractions. Also prepare a mixture of 30 mL of EtOAc and 170 mL of petroleum ether. To the 250 mL round bottom flask with the egg mixture, add 30 mL CH2Cl2, and stir thoroughly. Add the CH2Cl2 solution (not the egg mixture!) to the top of the column, and let the solvent go down on its own. When all the CH2Cl2 is down, rinse the inside top of the column with a little of the EtOAC/pet ether mixure. By this time, the solvent should have begun to drip out of the bottom of the column. Add more of the EtOAC/pet ether mixure to the top of the column, and apply gentle air pressure to the column as you collect 10 mL fractions. You should collect 15 fractions. Check the fractions by thin layer chromatography. Usually, the cholesterol comes in fractions 6-10. Rotovap the cholesterol fractions in a tared round bottom flask. You should be rewarded with iridescent rosettes of the product. Record the weight and the melting point.

               Option #1: Cholesterol forms a specific 2:1 complex with oxalic acid. Take up your crude cholesterol in 5 mL of 1,2-dichloroethane in a 50 mL Erlenmeyer flask. Add 80 mg of oxalic acid, and heat to reflux. Let the flask cool. After twenty minutes (clean up the lab!), swirl the flask in ice water. The contents should gel into a mush of crystals. Vacuum filter with 1,2-dichloroethane, suck dry, and spread out to dry. Take up the white residue in a 50 mL Erlenmeyer flask with 5 mL of water. Heat to reflux, chill in ice water, and vacuum filter. This should give pure white cholesterol, mp = 141-143 oC. Record the melting point and weight.

               Option #2: Take up the crude cholesterol in 5 mL of CH2Cl2 in a 50 mL round bottomflask. Spot a TLC plate in the left lane and the middle lane with the CH2Cl2 solution. Add 0.5 mL of Dess-Martin reagent in CH2Cl2, and warm briefly. Spot the solution in the middle and the right lanes of the same TLC plate, and develop in 1:4 EtOAC/pet ether. If a lot of starting material still remains, warm the solution again, and check it again. When the starting material is almost all converted, save a little of the CH2Cl2 solurion. Rotovap the rest of the solution, add 8 mL of acetone and 20 mg pTsOH, and warm it again to reflux. Spot the TLC plate middle and right, and develop it in 1:4 EtOAC/pet ether. This time, before you visualize the TLC plate with I2, check it with the UV light. You should see a UV spot for cholestenone. TLC Rf’s in 1:9 EtOAC/pet ether: cholesterol = 0.28, 5-cholesten-3-one = 0.60, 4-cholesten-3-one = 0.41. Alternatively, the oxidation can be carried out with the Brown protocol. To purify the 4-cholesten-3-one, add 2 g of flash silica gel to the acetone solution and rotavap it. Build a dry column with 10 grams of flash silica gel. Tap it to settle the silica gel, then add the dry powder with your reaction mixture, and then sand on top. Elute with a mixture of 2 mL of EtOAc and 100 mL of petroleum ether, then with a mixture of 5 mL of EtOAc and 95 mL of petroleum ether. Take 10 mL fractions, and check them by TLC. Evaporate the fractions containing 4-cholesten-3-one in a tared round bottom flask, and record the weight and melting point.